NEET Chemistry Question (2026) — Solution

Step-by-step solution

For the free expansion of an ideal gas, the external pressure is zero, so the work done W = 0. Since the process is isothermal at 300 K, the change in internal energy U = 0. From the first law of thermodynamics, q = U - W = 0. Since no heat is exchanged with the surroundings, the change in entropy of the surroundings is: S_ surroundings = q_ surr T = 0 Entropy is a state function, so the change in entropy of the system for an isothermal expansion is calculated using a reversible path between the same initial and final states: S_ system = nR ( V_2 V_1 ) Substituting n = 2 moles, V_1 = 10 L, and V_2 = 100 L: S_ system = 2R ( 100 10 ) = 2R (10) Using (10) = 2.303 _ 10 (10) = 2.303: S_ system = 2 2.303 R = 4.606 R Answer: S_ system =4.606 R; S_ surroundings =0